Let m be the mass of block. Initially, the block is in equilibrium and F is the up thrust on the block
F=T0+mg........(1)
The lift is accelerated upwards so g becomes (g + a)
Using Newton’s second law:
F′−T−mg=ma........(2)
From the figure, F′=F(g+ag)..........(3)
Solving equations (1),(2) and (3)
T=T0(1+ag)