Answer
$F = (1.18\times 10^{-14}~N)~\hat{i}+ (1.18\times 10^{-14}~N)~\hat{j}$
Work Step by Step
By symmetry, the gravitational forces on the central sphere from $m_1$ and $m_4$ cancel with each other.
$m_2 = 3.00~g$ and $m_3 = 1.00~g$ and the gravitational forces on the central sphere from these two masses are directed in opposite directions. We can replace these two masses with a single mass of $2.00~g$ at the location of $m_2$
We can find the distance between $m_2$ and $m_5$:
$r = \sqrt{(10.0~cm)^2+(10.0~cm)^2}$
$r = 14.14~cm$
We can find the magnitude of the net gravitational force on the central sphere:
$F = \frac{G~(2.00~g)(m_5)}{r^2}$
$F = \frac{(6.67\times 10^{-11}~N~m^2/kg^2)~(0.0020~kg)(0.0025~kg)}{(0.1414~m)^2}$
$F = 1.668\times 10^{-14}~N$
Note that this force is directed from the central sphere toward $m_2$
We can find the x-component of this force:
$F_x = (1.668\times 10^{-14}~N)~cos~45^{\circ}$
$F_x = 1.18\times 10^{-14}~N$
We can find the y-component of this force:
$F_x = (1.668\times 10^{-14}~N)~sin~45^{\circ}$
$F_x = 1.18\times 10^{-14}~N$
We can express the net gravitational force in unit-vector notation:
$F = (1.18\times 10^{-14}~N)~\hat{i}+ (1.18\times 10^{-14}~N)~\hat{j}$