Let X1,X2,s˙,Xn be n independent random variables each with mean 100 and standard deviation 30....

Step 1
${\displaystyle X_1,X_2,\dot{s},X_n}$ are independent ranodom variables with mean 100 and standard deviation 30
So, Var${\displaystyle \left(X_i\right)={30}^2=900}$
Now given that,
${\displaystyle X=\sum _i{X}_i}$
So ${\displaystyle E\left(X\right)=E\left(\sum _i{X}_i\right)=\sum _{i}E\left(X_i\right)=100n}$
and ${\displaystyle Var\left(X\right)=Var\left(\sum _i{X}_i\right)=\sum _{i}Var\left(X_i\right)}$ [ Covariances are zero due to independence]
So, Var(X) = 900n
Step 2
By Central Limit Theorem,
${\displaystyle \frac{X-\mu }{\sigma }\equiv \frac{X-100n}{\sqrt{900n}}\sim N(0,1)}$
Now, P(X>2000)>0,95
implies ${\displaystyle 1-P(X\le 2000)>0,95}$
${\displaystyle P(X\le 2000)<0,05}$
${\displaystyle P(\frac{X-100n}{\sqrt{900n}}\le \frac{2000-100n}{30\sqrt{n}})<0,05}$
implies ${\displaystyle \mathrm{\Phi }\left[\frac{2000-100n}{30\sqrt{n}}\right]<0,05=\mathrm{\Phi }(-1,645)}$
implies ${\displaystyle \frac{2000-100n}{30\sqrt{n}}<-1,645}$
Here you can solve this in two methods,
Trial and error method
Numerically using root finding method
Using trial and error method,
For n = 22
${\displaystyle \frac{2000-100n}{30\sqrt{n}}=-1,4213}$
For n=23
${\displaystyle \frac{2000-100n}{30\sqrt{n}}=-2,085}$
So, for n = 23 it is crossing -1.645. Hence n = 23