You have already worked out the mass of a tree if it has been growing continuously for the whole period t:

$\displaystyle W(t) = Q(1-e^{kt})^2$

If burns follow a poisson process it is known that the time between burns follows an $\displaystyle exp(\lambda)$ distribution.

Define $\displaystyle X_t$ as the amount of time that the tree has been continuously growing at time t. This is the time since the last burn (if there has been a burn), and t otherwise. I believe (but wont prove) that $\displaystyle X_t$ follows the following mixed distribution:

$\displaystyle X_t = r \text{ with probability density } \lambda e^{-\lambda r} ~for~ {0 \leq r \leq t$

$\displaystyle X_t = t \text{ with probability mass } e^{-\lambda t}$

it immediately follows that the Mass of the tree (Bt) is also a mixed distribution:

$\displaystyle B_t = W(r) \text{ with probability density } \lambda e^{-\lambda r} ~for~ {0 \leq r \leq t$

$\displaystyle B_t = W(t) \text{ with probability mass } e^{-\lambda t}$

Now evaluate E(B_t) in the normal way:

$\displaystyle E(B_t) = w(t)e^{-\lambda t} + \int_{0}^{t} w(r) \lambda e^{-\lambda r} dr $

$\displaystyle E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} + \int_{0}^{t} \left(Q(1-e^{-Kr})^2 \right) \lambda e^{-\lambda r} dr $

$\displaystyle E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} + Q \lambda \int_{0}^{t} (1-e^{-Kr})(1-e^{-Kr}) e^{-\lambda r} dr $

$\displaystyle E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} + Q \lambda \int_{0}^{t} \left( 1 - 2e^{-Kr} + e^{-2Kr} \right) e^{-\lambda r} dr $

$\displaystyle E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} + Q \lambda \int_{0}^{t} e^{-\lambda r} - 2e^{-r(k + \lambda)} + e^{-r(2K + \lambda)} dr $

$\displaystyle E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} + Q \lambda \left[ \frac{e^{-\lambda r}}{-\lambda} - \frac{2e^{-r(K + \lambda)}}{-K - \lambda} + \frac{e^{-r(2K + \lambda)}}{-2K - \lambda} \right]_{0}^{t} $

$\displaystyle E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} + Q \lambda \left[ \frac{2e^{-r(K + \lambda)}}{K + \lambda} - \frac{e^{-r(2K + \lambda)}}{2K + \lambda} -\frac{e^{-\lambda r}}{\lambda} \right]_{0}^{t} $

$\displaystyle E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} + Q \lambda \left( \frac{2e^{-t(K + \lambda)}}{K + \lambda} - \frac{e^{-t(2K + \lambda)}}{2K + \lambda} -\frac{e^{-\lambda t}}{\lambda} \right) - Q \lambda \left( \frac{2e^{-0(K + \lambda)}}{K + \lambda} - \frac{e^{-0(2K + \lambda)}}{2K + \lambda} -\frac{e^{-\lambda 0}}{\lambda} \right)$

$\displaystyle E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} + Q \lambda \left( \frac{2e^{-t(k + \lambda)}}{k + \lambda} - \frac{e^{-t(2K + \lambda)}}{2K + \lambda} -\frac{e^{-\lambda t}}{\lambda} \right) - Q \lambda \left( \frac{2}{K + \lambda} - \frac{1}{2K + \lambda} -\frac{1}{\lambda} \right)$

$\displaystyle E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} + Q \lambda \left( \frac{2(e^{-t(k + \lambda)}-1)}{k + \lambda} - \frac{e^{-t(2K + \lambda)}-1}{2K + \lambda} -\frac{e^{-\lambda t}-1}{\lambda} \right)$