Buck converter design problem/confusion
In a Buck Converter the inductor ripple current is determined by (approximately):
$$\dfrac{V_o.(1-V_o/V_i)}{L.f}$$

Which works out to be 6.66A for your values.

Designing the ripple current to be too high results in excessive RMS ripple losses in the inductor, output voltage ripple and bulk capacitor ripple losses (and heating). Too low a value results in a poor transient response limited by the inductor current slew rate.

A simple rule of thumb is to set the ripple current to 1/2 the max current of your design; for you that's 0.75A

From the formula you can see you only have a couple of options open to you.

Increase the switching frequency to about 800kHz which is very high for standard MOSFETs leading to excessive switching losses. Increase L to 133 \$\mu\$H (fairly large and higher DCR \$I^2R\$ losses); or a combination of both: i.e higher switching frequency and higher inductance. I would aim for about 300kHz and 47\$\mu\$H.