Answer
$5.3KN$
Work Step by Step
We can find the required weight as follows:
$F_b=\rho_{air}Vg$
$\implies F_b=\rho_{air}(\frac{4}{3}\pi r^3)g$
$F_b=(1.29)(\frac{4}{3}\pi(4.9)^3)(9.81)=6236N$
Now $W=F_b-m_{ballon}g-\rho_{He}V_{He}g$
We plug in the known values to obtain:
$W=6236-(3.2)(9.81)-(0.179)(\frac{4}{3}\pi (4.9)^3)(9.81)=5.3KN$
