CLT says that if each task duration $T_k$ has mean $\mu$ and standard deviation $\sigma$ then

$S_n = \dfrac 1 n \sum \limits_{k=1}^n T_k \to N\left(\mu, \dfrac{\sigma}{\sqrt{n}}\right)$

so we need to find $n \ni P\left[S_n < \dfrac {200}{n}\right] > 0.95$

This in turn means that

$\Phi\left(\dfrac{\frac{200}{n}-\mu}{\frac{\sigma}{\sqrt{n}}}\right)> 0.95$

if we let $p=\Phi^{-1}(0.95)$ then we can solve for $n$ as

$n = \left \lfloor \dfrac{400 \mu +p^2 \sigma ^2-\sqrt{p^4 \sigma ^4+800 \mu p^2 \sigma ^2}}{2 \mu ^2}\right \rfloor$

using $\mu=1.5,~\sigma=1,~p=1.645$

we get

$n = 121$