Answer
2003.
about $ 1.57$ million barrels/day
Work Step by Step
The graph of I(t) is a parabola that opens down (leading coefficient is negaative.)
$a=-0.015, b=0.1, c=1.4$
The vertex is the point of maximum, and has coordinates
$t=-\displaystyle \frac{b}{2a}$ and $I(-\displaystyle \frac{b}{2a})$
$t=-\displaystyle \frac{0.1}{-2\times 0.015}=\frac{10}{3}\approx 3.33$, which corresponds to year 2003.
$I(\displaystyle \frac{10}{3})\approx 1.57$ million barrels/day