Answer
$\phi = 1.03 \,\text{rad}$
Work Step by Step
The velocity of SHM is given by
\begin{align*}
v_s =- \omega x_{m} \sin (\omega t+\phi)
\end{align*}
At time $t$ = 0 the displacment is $x_s$ = 0.02 m, so get the displacement using the next equation
\begin{gather*}
x_s = x_{m} \cos (\omega (0)+\phi) \\
0.02 \,\text{m} = x_{m} \cos \phi \\
x_m = \dfrac{ 0.02 \,\text{m} }{\cos \phi }
\end{gather*}
For the velocity, $t$ = 0 and $v_s$ = 0.04 m/s will be
\begin{gather*}
v_s = - \omega x_{m} \sin (\omega t+\phi) \\
-0.04 \,\text{m/s} = - \omega x_{m} \sin \phi \\
x_m = \dfrac{ 0.04 \,\text{m/s} }{ \omega \sin \phi }
\end{gather*}
From both these equations we get the next
\begin{gather*}
\dfrac{ 0.02 \,\text{m} }{\cos \phi } = \dfrac{ 0.04 \,\text{m/s} }{ \omega \sin \phi } \\
\dfrac{ \sin \phi }{\cos \phi } = \dfrac{ 0.04 \,\text{m/s} }{ \omega (0.02 \,\text{m} ) } \\
\tan \phi = \dfrac{ 2 }{ \omega } \\
\phi = \tan^-1 \left( \dfrac{ 1 }{ \omega } \right)
\end{gather*}
Substitute by $\omega$ to get $\phi$
\begin{align*}
\phi = \tan^-1 \left( \dfrac{ 2}{ \omega } \right) = \tan^-1 \left( \dfrac{ 2}{ 1.2 \,\text{rad/s}} \right) = 59.03^o
\end{align*}
In rad, the phase constant is
$$\boxed{\phi = 1.03 \,\text{rad}}$$