From inspection the left hand side of this equation must be less than or equal to 1, so therefore $\displaystyle e^{0.02x} \le 0.05$, or $\displaystyle x \ge \frac {-ln(20)}{0.02}$. So the smallest root must be greater than 149.8. I think you'll find that the root is around 160.9.
Yes, the answer is near 150 (not exactly 150, but slightly more).
You can find the solution using a numerical technique, such as Newton's method, to find the root for $\displaystyle f(x) = e^{-0.02x}+0.95+ \sin(2x) = 0$. It involves first making a guess as to what the root is - call it $\displaystyle x_1$ - and finding the error of this guess:
Error = $\displaystyle e^{-0.02x_1}+0.95+ \sin(2x_1)$,
Then calculate the slope of f(x_1):
Slope = $\displaystyle -0.02 e^{-0.02x_1} + 2 \cos(2x_1)$
and set the next guess as $\displaystyle x_2 = x_1 - \frac {error}{slope}$. Iterate through this a few times and it will zero in on a value near 150.0007.