For BCC structure, the body diagonal = $4r = \sqrt{3}a$
$r = r^+ + r^- = 165 + 182 = 347$ pm
$\Rightarrow 4r = 1388$ pm
$a = \dfrac{4r}{\sqrt{3}} = \dfrac{1388}{1.732} \approx 800.7$ pm (this gives diagonal, now back-calculate $a$)
$a \approx \dfrac{4 \times 347}{\sqrt{3}} \Rightarrow a \approx 420$ pm