Let's derive the formula then we will compute the answer,
assuming mean = $m$,
$\Rightarrow \Large E\left ( X^2 \right ) = \sum_{0}^{\infty} x^2 P (x,m)$
$\Rightarrow \Large E\left ( X^2 \right ) = \sum_{0}^{\infty}( x(x-1) + x) \ P (x,m)$
$\Rightarrow \Large E\left ( X^2 \right ) = \sum_{0}^{\infty} x(x-1) P (x,m) + \sum_{0}^{\infty} x P (x,m)$
$\Rightarrow \Large E\left ( X^2 \right ) = \sum_{0}^{\infty} x(x-1) P (x,m) + m$ $∵ \ \sum_{0}^{\infty} x P (x,m) = m$
$\Rightarrow \Large E\left ( X^2 \right ) = \sum_{0}^{\infty} x(x-1) \frac{e^{-m} m ^x}{x!} + m$
$\Rightarrow \Large E\left ( X^2 \right ) = e^{-m}\sum_{0}^{\infty} x(x-1) \frac{ m ^x}{x!} + m$
$\Rightarrow \Large E\left ( X^2 \right ) = m^2 e^{-m} \left [\sum_{2}^{\infty} \frac{ m ^{x-2}}{\left (x-2 \right )!} \right ] + m$
$\Rightarrow \Large E\left ( X^2 \right ) = m^2 e^{-m} e^{m} + m$ $∵\left [\sum_{2}^{\infty} \frac{ m ^{x-2}}{\left (x-2 \right )!} \right ] = e ^m$
$\Rightarrow \Large E\left ( X^2 \right ) = m^2 + m$ .
Now given,
$\large E\left ( X^2 \right ) = 2$
$\large 2 = m^2 + m$
$\large m = 1$