LEDs are not resistors.

Ohms Law was originally written for metallic conductors, and other components that have an essentially constant resistance, in other words, a behaviour than can usefully be called resistance.

LEDs are more closely related to constant voltage devices, ie better modelled by constant voltage devices. If they conduct 20 mA at 1.85 V, they may well conduct uA at 1.75 V (very dim indeed), and amps at 1.95 V (gone up in smoke).

Your power supply voltage of 3.7 V suggests you are contemplating using a lithium ion battery with a nominal voltage of 3.7 V. In service, such a battery will deliver 4.2 down to 3ish volts, depending on where you stop discharging. Not a good power source for LEDs without current control.

To power LEDs successfully from such a source, use a LED plus resistor in series. With the nominal 3.7 and 1.85 V, you have 1.85 volts left across your current control resistor. Use a resistance of 1.85/20m = 93 ohms, or a few hundred ohms for controlling the current down to a friendlier 10 mA or so.

To address your question directly, you have 60 mA being drawn from a 3.7 V source, so the equivalent load resistance would be 3.7/0.06 = 62 ohms, which happens to be an E24 standard value.

LEDs are not resistors.

Ohms Law was originally written for metallic conductors, and other components that have an essentially constant resistance, in other words, a behaviour than can usefully be called a constant resistance.

LEDs are more closely related to constant voltage devices, ie better modelled by a constant voltage in series with a very small resistor. If they conduct 20 mA at 1.85 V, they may well conduct uA at 1.75 V (very dim indeed), and amps at 1.95 V (gone up in smoke).

Your power supply voltage of 3.7 V suggests you are contemplating using a lithium ion battery with a nominal voltage of 3.7 V. In service, such a battery will deliver 4.2 down to 3ish volts, depending on where you stop discharging. Not a good power source for LEDs without current control.

To power LEDs successfully from such a source, use a LED plus resistor in series. With the nominal 3.7 and 1.85 V, you have 1.85 volts left across your current control resistor. Use a resistance of 1.85/20m = 93 ohms, or a few hundred ohms for controlling the current down to a friendlier 10 mA or so.

The best way to power LEDs is with a constant current supply. A series resistor is a reasonable approximation to a constant current, but it wastes a lot of power. This is not a problem for a single 10 mA indicator LED on an instrument, but in a lighting installation of many watts, designers will specify switch mode constant current supplies to avoid wasting 50% of their input power on heat in resistors.

To address your question directly, you have 60 mA being drawn from a 3.7 V source, so the equivalent load resistance would be 3.7/0.06 = 62 ohms, which happens to be an E24 standard value.