Answer
$$P=190 \mathrm{\ kPa}$$
Work Step by Step
Knowing the standard air pressure value in several units allows us to set up a variety of
conversion factors:
$$P=\left(28 \mathrm{\ lb} / \mathrm{\ in} .^{2}\right)\left(\frac{1.01 \times 10^{5} \mathrm{\ Pa}}{14.7 \mathrm{\ lb} / \mathrm{in}^{2}}\right)=190 \mathrm{\ kPa}$$
