Step 1: The problem.
A particle in simple harmonic motion has a top speed $v$. The amplitude is made three times bigger and the time period is made twice as long. We want the new top speed.
Step 2: The top speed formula.
In SHM the maximum speed is at the centre and equals
\[ v_{\max} = \omega A \]
Since $\omega = \dfrac{2\pi}{T}$, this becomes
\[ v_{\max} = \frac{2\pi A}{T} \]
Step 3: Spot the dependence.
So the top speed grows with amplitude and shrinks with period:
\[ v_{\max} \propto \frac{A}{T} \]
Step 4: Note the changes.
New amplitude $A_2 = 3A$ and new period $T_2 = 2T$.
Step 5: Build the new speed.
\[ v' = \frac{2\pi (3A)}{2T} = \frac{3}{2}\cdot\frac{2\pi A}{T} \]
Step 6: Replace with v.
Since $\dfrac{2\pi A}{T} = v$:
\[ v' = \frac{3}{2}v = 1.5\,v \]
This is option (1).
\[ \boxed{v' = 1.5\,v} \]