Since $\displaystyle X_i$'s have mean $\displaystyle \mu$ and variance $\displaystyle \sigma^2$ for each $\displaystyle i$, then $\displaystyle \bar{X}$ has mean $\displaystyle \mu$ and variance $\displaystyle \frac{\sigma^2}{n}$. Consequently, the Chebyshev's inequality can be rewritten for $\displaystyle \bar{X}$ as:
$\displaystyle P(|\bar{X}-\mu|\geq a)\leq \frac{\sigma^2}{na^2}$
$\displaystyle =>P(\bar{X}\geq \mu+a, \bar{X}\leq \mu-a)\leq\frac{\sigma^2}{na^2}$ (breaking the modulus sign)
So $\displaystyle \frac{\sigma^2}{na^2}$ is the required upper limit.