$T = 2At\tau$

Where:

$T = 600 \, \text{N}\cdot\text{m} = 600\,000 \, \text{N}\cdot\text{mm}$

$A = 30(80) = 2400 \, \text{mm}^2$

$\tau = 80 \, \text{MPa}$

$600\,000 = 2(2400)(t)(80)$

$t = 1.5625 \, \text{ mm}$ *answer*

At any convenient center O within the section, the farthest side is the shorter one, thus, it is induced with the maximum allowable shear stress of 80 MPa.