As shown in figure, the tension in the horizontal cord is $30\text{N}$. The weight $W$ and tension in the string $OA$ in Newton are

The correct option is B $30\sqrt{3},60$
The FBD of the block:


The forces are:

As the block is in equilibrium, the summation of forces in any direction is zero:
$T\sin {30}^{\circ }=30N$
$\Rightarrow T=60N$

Again,
$T\cos {30}^{\circ }=W$
$W=60\times \frac{\sqrt{3}}{2}=30\sqrt{3}N$