Answer
$3.79\space m/s^{2}$
Work Step by Step
Let's apply Newton's second law in the vertical direction to the object.
$\uparrow F=ma$ ; Let's plug known values into this equation.
$F_{B}-mg=ma$
$V\rho_{c}g-V\rho_{h}g=V\rho_{h}a$
$a=\frac{(\rho_{c}-\rho_{h})g}{\rho_{h}}=\frac{(1.29\space kg/m^{3}-0.93\space kg/m^{3})(9.8\space m/s^{2})}{0.93\space kg/m^{3}}=3.79\space m/s^{2}$
Acceleration of the object in vertical direction = $3.79\space m/s^{2}$
