Given,

In right $\vartriangle ABC,\angle B={{90}^{\circ }}$

And, $BC=6cm,AB=8cm$

Let us consider, r be the radius of incircle with centre O and touches the sides AB, BC and CA at P, Q and R respectively.

as we can see, AR and AP are the tangents to the circle AR = AP

same above, $CR=CQ$ and $BQ=BP$

The radius of the circle are OP and OQ

$OP\bot AB$  and $OQ\bot BC$  and $\angle B={{90}^{\circ }}$  (given in the question)

Hence, BPOQ is a square

Thus, $BP=BQ=r$  (All four sides of a square are equal)

So,

$AR=AP=AB-PB$

$=8-r$

and $CR=CQ=BC-BQ$

$=6-r$

But $A{{C}^{2}}=A{{B}^{2+}}B{{C}^{2}}$  (Pythagoras Theorem applied)

$A{{C}^{2}}={{(8)}^{2}}+{{(6)}^{2}}$ $$

$A{{C}^{2}}=64+36$

$A{{C}^{2}}=100$

$A{{C}^{2}}={{(10)}^{2}}$

So, $AC=10cm$

$\Rightarrow AR+CR=10$

$\Rightarrow 8-r+6-r=10$

$\Rightarrow 14-2r=10$

$\Rightarrow 2r=14-10$

$\Rightarrow 2r=4$

$\Rightarrow r=2$