One light year $=$ speed of light $\times$ one year
or $1 ly=3\times 10^8\times (24\times 3600)=94608\times 10^{11} m$
So, $4.29 ly=4.29\times (94608\times 10^{11})=4.058\times 10^{16} m$
As $1 $ parsec $=3.08\times 10^{16} m$
(Parsec is a unit of length used to measure large distances to objects outside our Solar System)
Thus, $4.29 ly=\dfrac{4.058\times 10^{16}}{3.08\times 10^{16}}=1.32$ parsec
Now angular displacement , $\theta=\dfrac{d}{D}$
where $d=$ diameter of earth's orbit $= 3\times 10^{11} m$ and
$D=$ distance of star from the earth $=4.058\times 10^{16} m$
So, $\theta=\dfrac{3\times 10^{11}}{4.058\times 10^{16}}=7.39\times 10^{-6}$ rad
As $1 sec=4.85\times 10^{-6} rad$ so, $7.39\times 10^{-6} rad=\dfrac{7.39\times 10^{-6}}{4.85\times 10^{-6}}=1.52 sec$