Let the number of students be $n$.

The original mean is $10$. When one student's marks increase from $8$ to $12$, the total marks increase by $12-8=4.$

The new mean becomes $10.2$, so the increase in mean is $10.2-10=0.2.$

Hence, $\frac{4}{n}=0.2$ which gives $n=\frac{4}{0.2}=20.$

Use the variance formula to find the original sum of squares.

Original variance: $\sigma^2=4$ and mean $\mu=10$.

Using $\sigma^2=\frac{\sum x_i^2}{n}-\mu^2,$ we get $4=\frac{\sum x_i^2}{20}-10^2.$

So, $4=\frac{\sum x_i^2}{20}-100$ $\frac{\sum x_i^2}{20}=104$ $\sum x_i^2=2080.$

Update the sum of squares after changing one mark from $8$ to $12$.

The old contribution was $8^2=64,$ and the new contribution is $12^2=144.$

So the sum of squares increases by $144-64=80.$

Therefore, new sum of squares is $2080+80=2160.$

Find the new variance.

New mean is $10.2$, so $\text{New variance}=\frac{2160}{20}-(10.2)^2.$

Now, $\frac{2160}{20}=108$ and $(10.2)^2=104.04.$

Hence, $\text{New variance}=108-104.04=3.96.$

Match with the options.

$3.96$ corresponds to Option C.