A cavity of radius $\frac{R}{4}$ is made inside a non-conducting solid sphere as shown in the figure.







Let $Q_0$ be the charge, present on the sphere when cavity was absent.
Hence, $E=E_{Q_0}-E_{\text {cavity }}$
Where, $E=$ electric field at centre of cavity,
$E_{Q_0}=$ electric field at a distance $\frac{R}{2}$ from centre and
$E_{\text {cavity }}=$ electric field due to cavity itself at its centre.
Hence, $E=\frac{Q_0}{4 \pi \varepsilon_0 R^3}\left(\frac{R}{2}\right)-0$
$\Rightarrow \quad E=\frac{Q_0}{4 \pi \varepsilon_0 R^2}\left(\frac{1}{2}\right)$


Now, it is given that $Q$ be the charge after the creation of the cavity.
So,
$$
\frac{Q}{Q_0}=\frac{\frac{4}{3} \pi\left(R^3-\left(\frac{R}{4}\right)^3\right)}{\frac{4}{3} \pi R^3}
$$
$\because$ Since, for a uniform charged density object, charge $\propto$ volume
$$
\begin{array}{ll}
\Rightarrow & \frac{Q}{Q_0}=\frac{63}{64} \\
\Rightarrow & Q_0=\frac{Q 64}{63}
\end{array}
$$
So, from the Eq. (i) and (ii), we get
$$
E=\frac{Q}{4 \pi \varepsilon_0 R^2}\left(\frac{64}{63} \times \frac{1}{2}\right)
$$

$$
\begin{aligned}
& \qquad=0.5079\left[\frac{Q}{4 \pi \varepsilon_0 R^2}\right) \\
& \Rightarrow \quad E=0.51\left[\frac{Q}{4 \pi \varepsilon_0 R^2}\right] \\
& \text { Hence, } K=0.51 \\
& \therefore \text { So, the correct option is }(\mathrm{3}) .
\end{aligned}
$$