Step 1: Use the heat balance idea.
When hot and cold water mix, the heat that leaves the hot water is exactly the heat that enters the cold water. Nothing is lost to the outside. This is the principle of calorimetry, and it is what lets us find the final temperature.
Step 2: Note that both are water.
Heat is $Q=mc\,\Delta T$. Since both samples are water, the specific heat $c$ is the same on both sides and simply cancels. So we only need the masses and temperature changes.
Step 3: Write the two amounts.
Cold water has mass $100\,\text{g}$ at $60^\circ C$. Hot water has mass $180\,\text{g}$ at $95^\circ C$. Let the final shared temperature be $T$.
Step 4: Form the balance equation.
Heat gained by cold equals heat lost by hot. \[ 100\,(T-60)=180\,(95-T) \]
Step 5: Open the brackets.
Multiply out both sides. \[ 100T-6000=17100-180T \] Now bring the $T$ terms together: $100T+180T=17100+6000$, which is $280T=23100$.
Step 6: Solve for $T$.
Divide both sides by $280$. \[ T=\frac{23100}{280}=82.5^\circ C \] This sits between $60^\circ C$ and $95^\circ C$, so it is sensible. It matches option 2. \[ \boxed{82.5^\circ C} \]