$\overrightarrow{\mathrm{b}}-2\overrightarrow{\mathrm{c}}=\mathrm{\lambda }\overrightarrow{\mathrm{a}}$
$\Rightarrow {\left|\overrightarrow{\mathrm{b}}-2\overrightarrow{\mathrm{c}}\right|}^2=\mathrm{\lambda }{\left|\overrightarrow{\mathrm{a}}\right|}^2$
$\Rightarrow {\left|\overrightarrow{\mathrm{b}}\right|}^2+4{\left|\overrightarrow{\mathrm{c}}\right|}^2-4\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}={\mathrm{\lambda }}^{2}2$
$$\begin{gathered} \Rightarrow 16+4-4\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}=2{\mathrm{\lambda }}^2 \\ \Rightarrow {\mathrm{\lambda }}^2=10-2\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}} ... \left(1\right) \end{gathered}$$
Now, ${\left|\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}\right|}^2+{\left|\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}\right|}^2={\left|\overrightarrow{\mathrm{b}}\right|}^2{\left|\overrightarrow{\mathrm{c}}\right|}^2$
$15+{\left|\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}\right|}^2=16$
$\Rightarrow \left|\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}\right|=\pm 1$
The angle between $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ is acute
Hence, $\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}=1$
from $\left(1\right)$

${\mathrm{\lambda }}^2=8\Rightarrow \mathrm{\lambda }=\pm 2\sqrt{2}$