Answer
$2.2Kg$
Work Step by Step
We can find the required mass as follows:
$\Sigma F_x=-F+T_1cos30^{\circ}+T_2cos30^{\circ}=0$
$\implies \Sigma F_x=-F+2mgcos30^{\circ}$ (As $T_1=T_2=mg$)
This can be rearranged as:
$m=\frac{F}{2gcos30^{\circ}}$
We plug in the known values to obtain:
$m=\frac{37}{2(9.81)cos30^{\circ}}$
$m=2.2Kg$
