At the top of the circle, the net force on the ball is pointing downward, so that by Newton's second law
[tex]\sum F = -ma \implies F_{\rm tension} + F_{\rm weight} = \dfrac{mv^2}R[/tex]
where [tex]a=\frac{v^2}R[/tex] because the ball is undergoing circular motion with constant speed.
Plug in everything you know and solve for [tex]v[/tex].
[tex]4.0\,\mathrm N + (0.17\,\mathrm{kg}) g = \dfrac{(0.17\,\mathrm{kg})v^2}{0.62\,\rm m} \\\\ \implies v^2 = \dfrac{0.62\,\rm m}{0.17\,\rm kg} \left(4.0\,\mathrm N + (0.17\,\mathrm{kg}) g\right) \\\\ \implies v^2 \approx 21 \dfrac{\mathrm m^2}{\mathrm s^2} \implies \boxed{v \approx 4.5\dfrac{\rm m}{\rm s}}[/tex]
Plug this into the acceleration equation and solve for [tex]a[/tex].
[tex]a = \dfrac{21\frac{\mathrm m^2}{\mathrm s^2}}{0.62\,\rm m} \implies \boxed{a \approx 33\dfrac{\rm m}{\mathrm s^2}}[/tex]
