(d) The given situation is shown in the following figure,
The acceleration given to wedge to the left is a. The block has a pseudo acceleration to the right, pressing against the wedge because of which the block does not move. In this case, ∴‌mg‌sin‌θ=ma‌cos‌θ ⇒‌a=‌
g‌sin‌θ
cos‌θ
‌‌......(i) Force exerted by the wedge on the block is equal to normal reaction on the block, i.e. R=mg‌cos‌θ+ma‌sin‌θ =mg‌cos‌θ+m⋅g‌