Answer
$0.914 \;\rm m$
Work Step by Step
We need to find the forces exerted on the board, as seen in the second figure below. The board is assumed to be uniform so its center of mass is just in the middle as we drew below.
Now we know that this board is in equilibrium which means that the net torque exerted on it is zero.
Thus, the net torque around the pivot point is zero as well.
$$\sum \tau_{pivot}=F_{n,pivot}(0)+m_{\rm girl}gx_{cm}+m_{\rm board}g(1.25)-F_{n,scale}(2.5)=0$$
We chose counterclockwise around the pivot point to be the positive direction.
$$ m_{\rm girl}gx_{cm}+1.25m_{\rm board}g -2.5F_{n,scale}=0$$
Solving for the center of mass of the girl,
$$ x_{cm} =\dfrac{2.5F_{n,scale} -1.25m_{\rm board}g}{m_{\rm girl}g}$$
Noting that the normal force exerted by is given by $25 g$
$$ x_{cm} =\dfrac{2.5(25\color{red}{\bf\not} g)-1.25m_{\rm board}\color{red}{\bf\not} g}{m_{\rm girl}\color{red}{\bf\not} g}$$
Hence,
$$ x_{cm} =\dfrac{[2.5\times 25]-[1.25\times6.1]}{ 60}$$
$$ x_{cm} =\color{red}{\bf 0.914}\;\rm m$$
