Answer
$v_2 = 4.0~m/s$
Work Step by Step
Since the string pulls at a $90^{\circ}$ angle to the motion of the mass, there is no net external torque on the system. Therefore angular momentum is conserved:
$I_2~\omega_2 = I_1~\omega_1$
$(m~r_2^2)(\frac{v_2}{r_2}) = (m~r_1^2)(\frac{v_1}{r_1})$
$(r_2)(v_2) = (r_1)(v_1)$
$v_2 = \frac{ (r_1)(v_1)}{v_2} = \frac{(0.80~m)(2.4~m/s)}{0.48~m}$
$v_2 = 4.0~m/s$