Step 1:

The given corner points are $(0,0),(0,40),(20,40),(60,20),(60,0)$

The objective function is $Z=4x+3y$

For the points $(x,y)$ the objective function subject to $Z=4x+3y$

Step 2:

At $(0,0)$ the objective function $z=4x+3y\Rightarrow z=0$

At $(0,40)$ the objective function $z=4x+3y\Rightarrow Z=4\times 0+3\times 40=120$

At $(20,40)$ the objective function $z=4x+3y\Rightarrow 4\times 20+3\times 40=200$

At $(60,20)$ the objective function $z=4x+3y\Rightarrow 4\times 60+3\times 20=300$

At $(60,0)$ the objective function $z=4x+3y\Rightarrow 4\times 60+3\times 0=240$

Step 3:

Hence the maximum of $Z$ is 300

For column B the maximum is 325.

Clearly quantity of column B is greater.

Hence the correct option is $B$.