Step 1: Let $O$ be the common centre. Let $AB$ be the chord of the larger circle (radius $OA=10$ cm) touching the smaller circle (radius 6 cm) at $P$.
Step 2: Since $AB$ is a tangent to the smaller circle at $P$, the radius $OP$ is perpendicular to $AB$.
$$OP \perp AB,\qquad OP=6\text{ cm}$$
Step 3: A perpendicular from the centre to a chord bisects the chord, so $AP=PB=\dfrac{1}{2}AB$.
Step 4: Apply Pythagoras theorem in right $\triangle OPA$.
$$OA^{2}=OP^{2}+AP^{2}$$
$$10^{2}=6^{2}+AP^{2}$$
Step 5: Solve for $AP$.
$$AP^{2}=100-36=64 \;\Rightarrow\; AP=8\text{ cm}$$
Step 6: Length of the chord
$$AB=2\times AP=2\times 8=16\text{ cm}$$
Hence the required chord is 16 cm long.