Okay, well it looks like you'll have to calculate the mean and standard deviation for the data first:
\[\begin{align*}\mu&=E(X)\\&=\sum_{x=0}^5x f(x)\\&=2.72\\\\
\sigma^2&=V(X)\\
&=E(X^2)-\bigg[E(X)\bigg]^2\\
&=\sum_{x=0}^5x^2 f(x)-2.72^2\\
&=9.62-7.4\\
&=2.22\\
\sigma&=\sqrt{2.22}\approx1.49
\end{align*}\]
Ok, so now you want to find \(X=x\) such that, according to the inequality,
\[P(|X-\mu|\ge k\sigma)\le\color{red}{\frac{1}{k^2}}\]
The red value is given to be \(0.10\), so
\[\frac{1}{k^2}=0.10~~\Rightarrow~~k=3.16\]
\[|X-2.72|\ge 3.16\cdot1.49\]
Solving for \(X\), it looks the manager should upgrade to \(x\approx7.4284\approx8\) slots for customers, i.e. increase the current capacity by 6.