With 10 integers available, [tex]X[/tex] has PMF
[tex]\mathbb P(X=x)=\begin{cases}\dfrac1{10}&\text{for }0\le x\le9,x\in\mathbb Z\\\\0&\text{otherwise}\end{cases}[/tex]
We're interested in the statistics of the new random variable [tex]Y=5X[/tex]. To do this, we need to know the PMF for [tex]Y[/tex]. This isn't too hard to find.
[tex]\mathbb P(Y=y)=\mathbb P(5X=y)=\mathbb P\left(X=\dfrac y5\right)[/tex]
Since the PMF for [tex]X[/tex] gives a value of [tex]\dfrac1{10}[/tex] whenever [tex]x[/tex] is an integer between 0 and 9, it follows that [tex]\dfrac y5[/tex] must also be an integer for the PMF to give the identical value of [tex]\dfrac1{10}[/tex]. This means
[tex]\mathbb P(Y=y)=\begin{cases}\dfrac1{10}&\text{for }y\in\{0,5,\ldots,45\}\\\\0&\text{otherwise}\end{cases}[/tex]
Now the mean (expectation) is
[tex]\mathbb E(Y)=\displaystyle\sum_yy\mathbb P(Y=y)=\frac1{10}\sum_{y\in\{0,\ldots,45\}}y[/tex]
[tex]\mathbb E(Y)=\dfrac{225}{10}=22.5[/tex]
The variance would be
[tex]\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2[/tex]
[tex]\mathbb V(X)=\displaystyle\sum_yy^2\mathbb P(Y=y)-\mathbb E(Y)^2[/tex]
[tex]\mathbb V(X)=\displaystyle\frac1{10}\sum_{y\in\{0,\ldots,45\}}y^2-\left(\frac{225}{10}\right)^2[/tex]
[tex]\mathbb V(X)=\dfrac{7125}{10}-\dfrac{50625}{100}=206.25[/tex]
The standard deviation is the square root of the variance, so you have
[tex]\sqrt{\mathbb V(X)}=\sqrt{206.25}\approx14.3614[/tex]
