Answer
$F=340N$
Work Step by Step
Note that the minimum amount of force needed to overcome static friction is $$F_{s,max}=\mu_smg$$ Substituting known values of $\mu_s=0.50$, $m=68kg$, and $g=9.8m/s^2$ yields a minimum force of $$F_{s,max}=(0.50)(68kg)(9.8m/s^2)$$ $$F=330N$$ This is the horizontal component of the force, however. By drawing a triangle and noting that the force applied is 15 degrees above the horizontal, the force needed is actually equal to $\frac{F}{cos\theta}$. Substituting known values of $F=330N$ and $\theta=15^{\circ}$ yields a needed force of $$F_{min}=\frac{330N}{cos(15^{\circ})}=340N$$
