Answer
$$1.25T_0$$
Work Step by Step
Let $m$ and $V$ be the mass and volume of the block respectively.
When the container is at rest:
$T_0+mg=F_b$
$T_0=F_b-mg$
Where, $F_b$ is the buoyant force acting on the block.
$T_0=V\rho g-mg$
When the container is given an upward acceleration of $a=0.250g$, the apparent weight of the block becomes $W^{'}=m(a+g)$ and the buoyant force yields $F_b^{'}=\rho V(a+g)$.
The tension $(T_1)$ in the string becomes:
$T_1=V\rho (a+g)-m(g+a)$
or, $T_1=V\rho a+V\rho g-mg-ma$
or, $T_1=(V\rho g-mg)+(V\rho g-mg)\frac{a}{g}$
or, $T_1=(V\rho g-mg)(1+\frac{a}{g})$
or, $T_1=T_0(1+\frac{a}{g})$
or, $T_1=T_0(1+\frac{0.250g}{g})$
or, $\boxed{T_1=1.25T_0}$
