Let the five observations are $x_1, x_2, x_3, x_4, x_5$.
So, mean $=\frac{x_1+x_2+x_3+x_4+x_5}{5}=9$
(given)
$\Rightarrow \quad x_1+x_2+x_3+x_4+x_5=45$
and standard deviation $=\sum_{i=1}^5 \sqrt{\frac{\left(9-x_i\right)^2}{5}}=0$ (given)
$\Rightarrow \quad \sum_{i=1}^5\left(9-x_i\right)^2=0 \Rightarrow x_i=9 \forall i$
Now, let observation $x_5$ is changed with $y$, so
$$
\begin{array}{lrr}
\text { mean }=\frac{x_1+x_2+x_3+x_4+y}{5}=10 & \text { (given) } \\
\Rightarrow & x_1+x_2+x_3+x_4+y=50 & \\
\Rightarrow & 45-x_5+y=50 & \text { (from Eq. (i)) } \\
\Rightarrow & x_5+5=y \Rightarrow y=14 & \left\{\because x_5=9\right\}
\end{array}
$$
(given)

Now, the changed standard deviation is
$$
\begin{aligned}
& \sqrt{\frac{\left(10-x_1\right)^2+\left(10-x_2\right)^2+\left(10-x_3\right)^2}{\frac{+\left(10-x_4\right)^2+(10-y)^2}{5}}} \\
= & \sqrt{\frac{1+1+1+1+16}{5}}=\sqrt{\frac{20}{5}}=\sqrt{4}=2
\end{aligned}
$$