We have a gaussian distribution
$$ X \sim N(\mu,\sigma^2)$$
where $\mu = 4$ and $\sigma^2 =1.5$ .
Probability is given by :
$P(x<c)=0.35$
$c$ needs to be calculated.
And we got
$$z\left(\frac{c-4}{1.5}\right)=0.35$$
$$\frac{c-4}{1.5}=-0.385$$
$$c=3.4225$$
Do I get $c$ from $\dfrac{1}{\sqrt{2\pi}}e^{-0.5\left(\dfrac{x-\mu}{\sigma}\right)^2}$?
If $\sigma = 1.5$, then $P\{X < c\} = P\{\frac{X-\mu}{\sigma} < \frac{c-\mu}{\sigma}\} = P\{Z < \frac{c-4}{1.5}\} = 0.35,$ where $Z$ is standard normal.
Then from normal tables $(c - 4)/1.5 \approx -0.3853.$ Solve for $c.$ If $\sigma^2 = 1.5$ as you say, then $\sigma = 1.224745$ and adjust the denominator accordingly before solving.
If $\sigma = 1.5$, then the solution is approximately 3.422. If $\sigma^2 = 1.5$, then the solution is approximately 3.528.