Given,

$f:\mathrm{ℝ}\to \mathrm{ℝ}$ be a function defined by $f\left(x\right)={\log }_{\sqrt{m}} \left\{\sqrt{2}\left(\sin x-\cos x\right)+m-2\right\}$, for some $m$,

Also given the range of $f$ is $\left[0,2\right]$,

Now we know that,

$-\sqrt{2}\le \sin x-\cos x\le \sqrt{2}$

$\Rightarrow -2\le \sqrt{2}\left(\sin x-\cos x\right)\le 2$

(Assuming $\sqrt{2}\left(\sin x-\cos x\right)=k$)

$\Rightarrow -2\le k\le 2 \dots \left(1\right)$

Now taking function $f\left(x\right)={\log }_{\sqrt{\mathrm{m}}}\left(k+m-2\right)$

Given, $0\le f\left(x\right)\le 2$

$\Rightarrow 0\le {\log }_{\sqrt{m}}\left(k+m-2\right)\le 2$

$\Rightarrow 1\le k+m-2\le m$

$\Rightarrow -m+3\le k\le 2 \dots \left(2\right)$

Now from equations $\left(1\right) \& \left(2\right)$, we get

$-m+3=-2$

$\Rightarrow \mathrm{m}=5$