For a certain solid isotropic material, the modulus of rigidity is...

Let the bulk modulus be $K$ and the modulus of rigidity be $\eta$.

Given that $\eta = 0.75 K = \dfrac{3}{4} K$.

The standard relation connecting Poisson's ratio $\sigma$, bulk modulus $K$, and modulus of rigidity $\eta$ is:
$\sigma = \dfrac{3K - 2\eta}{6K + 2\eta}$

Substitute $\eta = 0.75 K$ into the formula:
$\sigma = \dfrac{3K - 2(0.75 K)}{6K + 2(0.75 K)}$

$\sigma = \dfrac{3K - 1.5K}{6K + 1.5K}$

$\sigma = \dfrac{1.5K}{7.5K}$

$\sigma = \dfrac{1.5}{7.5} = \dfrac{1}{5} = 0.2$

Alternatively, using the relations $Y = 3K(1 - 2\sigma)$ and $Y = 2\eta(1 + \sigma)$:
$3K(1 - 2\sigma) = 2(0.75 K)(1 + \sigma)$
$3(1 - 2\sigma) = 1.5(1 + \sigma)$
$2(1 - 2\sigma) = 1 + \sigma$
$2 - 4\sigma = 1 + \sigma$
$5\sigma = 1 \Rightarrow \sigma = 0.2$

Answer: $0.2$